Answers
[tex]\frac{(x+3)^2}{x-5} \div \frac{x^2 - 9}{x^2 - 8x +15} \\ = \frac{x^2 + 9}{x-5} \div \frac{x^2 - 9}{x^2 - 8x +15} \\ = \frac{x + 9}{-5} \div \frac{-9}{-8x + 15} \\ = \frac{x}{-5} \div \frac{-1}{-8x + 15} \\ = - \frac{x(8x-15)}{5}[/tex]
Note : You only need to substitute, divide and eliminate the numbers to simplify.
Hope it helps.
ANSWER DETAILSSubject : Mathematics
Class : 9th Grade
Chapter : -
Categorization Code : -
Keywords : Substitution, Algebra.
Answer:
[tex] \frac{x + 3}{x - 5} [/tex]
Step-by-step explanation:
The first order of business is to get the denominator the same on both fractions. To do this change the 2nd terms denominator into intercept form. See below.
[tex] {x}^{2} - 8x + 15 = (x - 5)(x - 3)[/tex]
Do this for the numerator on the 2nd term also.
[tex] {x}^{2} - 9 = (x - 3)(x + 3)[/tex]
Now we can exclude the (x-3) from both the numerator and denominator of the second term. Now we have the equation below.
[tex] \frac{(x + 3 {)}^{2} }{x - 5} \div \frac{ x + 3}{x - 5} [/tex]
Now the denominators are the same. Now we need to exclude x+3 from both numerators. Notice that in the firt term, you do not exclude both x+3, only 1.
[tex] \frac{x + 3}{x - 5} [/tex]
There is your answer
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